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Lady bird [3.3K]
3 years ago
9

YALL PLEASE PLEASE HELP IF YOU KNOE THE ANSWER TO BOTH OF THEM IT WILL BE VERY NICE PLEASE HELP

Physics
1 answer:
Juliette [100K]3 years ago
8 0

Answer:

2.5m/s²

Force  = mass x acceleration  

Explanation:

Given parameters

Mass  = 200kg

Force  = 500N

Unknown:

Acceleration  = ?

Solution:

According to the Newton's second law of motion "the constant force applied on a massive body causes it to accelerate".

 Mathematically;

    Force  = mass x acceleration  

Now insert the parameters and solve;

   Acceleration  = \frac{force}{mass}  

   Acceleration = \frac{500}{200}   = 2.5m/s²

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Answer:

He is known as the first microbiologist and also “the Father of Microbiology” because he was the first to observe bacteria underneath a microscope. He made many other significant discoveries in the field of biology and also made important changes to the microscope.

Explanation:

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Which statement is true about acceleration?<br> who is an underated physicist?
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4 years ago
Which of the following is most closely related to the specific heat of a substance?A. its temperatureB. its melting pointC. its
swat32

Answer:

option (C)

Explanation:

The amount of heat required to raise the temperature of substance of unit mass by unit degree is called specific heat of that substance.

Its SI unit is Joule / Kg °C.

Every material has a constant value of specific heat.

So, option (c) is correct.

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3 years ago
An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

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3 years ago
A control clinic offers a program that guarantees a weight loss of up to 0.46 kg in one week. Express the weight loss in a ratio
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Answer:

0.76 mg/s

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